Let $(X, d)$ be a metric space. For $f:X\to X$, $x\in X$ and $c>0$, take $\Gamma_c(x, f)= \{y: d(x_n, y_n)<c, \forall n\in\mathbb{Z}\}$ where $\{x_n\}_{n\in\mathbb{Z}}$ and $\{y_n\}_{n\in\mathbb{Z}}$ are orbits of $x, y$, respectively, this means that $x_{n+1}= f(x_n)$ with $x_0=x$ and $f(y_{n})=y_{n+1}$ with $y_0=y$.
$f:X\to X$ is called expansive map/ homeomorphism, if there is $c>0$ such that $\Gamma_c(x, f)=\{x\}$ for all $x\in X$.
It is known that there is no expansive homeomorphism on one-dimensional compact manifold. This means that there is no expansive homeomorphism on circle and interval. But $f:S^1\to S^1$ with $z\to 2z$, is an expansive local homeomorphism. Thus there is expansive local homeomorphism on circle.
Is there an expansive local homeomorphism on $[0, 1]$?
Please help me to know it.
The answer is no. If $f : [0,1] \to [0,1]$ is a local homeomorphism, then it must locally increasing, so just increasing since $[0,1]$ is connected. But this means that it is in fact a homeomorphism of $[0,1]$.
In particular, it cannnot be expansive.